Class 11 RD Sharma Solutions - Chapter 12 Mathematical Induction - Exercise 12.2 | Set 3

Exercise 12.2 | Set 3 of Chapter 12 in RD Sharma's Class 11 mathematics textbook presents an advanced exploration of Mathematical Induction. This set offers students a collection of challenging problems that push the boundaries of their inductive reasoning skills. It includes complex algebraic identities, intricate inequalities, and sophisticated number theory problems. By engaging with these advanced exercises, students further refine their proof-writing abilities, deepen their understanding of mathematical patterns, and develop a more nuanced approach to problem-solving. This set aims to prepare students for high-level mathematical reasoning encountered in advanced mathematics courses and competitive examinations.

What is Mathematical Induction?

Mathematical Induction is a proof technique used to establish that a given statement is true for all natural numbers or a specified subset of them. It consists of two main steps: the base case and the inductive step. In the base case, we prove the statement for the smallest value in the set (typically 1 or 0). The inductive step involves assuming the statement is true for an arbitrary value k (the inductive hypothesis) and then proving it holds for k+1. If both steps are successfully proven, we can conclude that the statement is true for all natural numbers greater than or equal to the base case, leveraging the well-ordering principle of natural numbers.

Question 33. Prove that n¹¹/11 + n⁵/5 + n³/3 - 62/165n is true for all n ∈ N.

Solution:

Let, P(n) = n¹¹/11 + n⁵/5 + n³/3 - 62/165n

Step 1:

Now, let us check P(n) for n = 1.

So, P(1) = 1/11 + 1/5 + 1/3 - 62/165 = 1

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m,

Let, P(m) = m¹¹/11 + m⁵/5 + m³/3 - 62/165m

So, m¹¹/11 + m⁵/5 + m³/3 - 62/165m = λ, where λ ∈ N is a positive integer.

Step 3:

Now, we have to prove for P(m + 1) is true.

P(m + 1) = (m + 1)¹¹/11 + (m + 1)⁵/5 + (m + 1)³/3 - 62/165(m + 1)

= 1/11(m¹¹ + 11m¹⁰ + 55m⁹ + 165m⁸ + 330m⁷ + 462m⁶ + 462m⁵ + 330m⁴ + 

165m³ + 55m² + 11m + 1) + 1/5(m⁵ + 5m⁴ + 10m³ + 10m² + 5m + 1) + 

1/3(m³ + 3m² + 3m + 1) + 62/165(m + 1)

= λ + m⁶ + 3m⁵ + 5m⁴ + 5m³ + 3m² + m + m⁴ + 2m³ + 2m² + m + m² + m + m + 1

As λ is positive, so it is a positive integer.

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 34. Prove that (1/2)tan(x/2) + (1/4)tan(x/4) +.....+ (1/2ⁿ)tan(x/2ⁿ) = (1/2ⁿ)cot(x/2ⁿ) - cotx for all n ∈ N and 0< x < π/2.

 Solution:

Let, P(n) = (1/2)tan(x/2) + (1/4)tan(x/4) +.....+ (1/2ⁿ)tan(x/2ⁿ) = (1/2ⁿ)cot(x/2ⁿ) - cotx, for all n ∈ N and 0< x < π/2.

Step 1:

Now, let us check P(n) for n = 1.

LHS = 1/2 tan(x/2)

RHS = (1/2)cot(x/2) - cot(x) = 1/(2tan(x/2)) - 1 tan(x)

= 1/(2tan(x/2)) - 1/(2tan(x/2))/(1 - tan²(x/2))

= 1/(2tan(x/2)) - (1 - tan²(x/2))/(2tan(x/2))

=(1/2)tan(x/2)

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m,

P(m) = (1/2)tan(x/2) + (1/4)tan(x/4) +.....+ (1/2^m)tan(x/2^m) = (1/2^m)cot(x/2^m) - cotx

Step 3:

Now, we have to prove for P(m + 1) is true.

P(m + 1) = (1/2)tan(x/2) + (1/4)tan(x/4) +.....+  (1/2^m)tan(x/2^m) + (1/2^{(m + 1)})tan(x/2^{(m + 1)}) = (1/2^{(m + 1)})cot(x/2^{(m + 1)}) - cotx

So, = (1/2^m)cot(x/2^m) - cotx + (1/2^(m+1))cot(x/2^(m+1))

= (1/2^m)cot (x/2^m) + (1/2^(m+1))tan(x/2^(m+1)) - cotx

= 1/(2^mtan(2x/2^(m+1)) + (1/2^(m+1))tan(x/2^(m+1)) - cotx

= [(1 - tan²(x/2^(m+1)))/2^(m+1).tan(x/2^(m+1))] + (1/2^(m+1))tan(x/2^(m+1)) - cotx

= (1/2^(m+1))cot(x/2^(m+1)) - cotx

Now,

(1/2)tan(x/2) + (1/4)tan(x/4)+.....+ (1/2^m)tan(x/2^m) + (1/2^(m+1))tan(x/2^(m+1)) = (1/2^(m+1))cot(x/2^(m+1)) - cotx

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 35. Prove that (1 - 1/2²)(1 - 1/3²)(1 - 1/4²)......(1 - 1/n²) = (n + 1)/2n is true for all n ∈ N.

Solution:

Let, P(n) =  (1 - 1/2²)(1 - 1/3²)(1 - 1/4²)......(1 - 1/n²) = (n + 1)/2n

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1 - 1/2² = (2 + 1)/2.2

or, 3/4 = 3/4

So, P(2) is true.

Step 2:

Let us consider P (n) be the true for n = k, So, P(k) is

P(k) =  (1 - 1/2²)(1 - 1/3²)(1 - 1/4²)......(1 - 1/k²) = (k + 1)/2k

Step 3:

Now, we have to prove for P(k + 1) is true. i.e. 

P(k + 1) = (1 - 1/2²)(1 - 1/3²)(1 - 1/4²)......(1 - 1/(k + 1)²) = (k + 2)/2k

Now, (1 - 1/2²)(1 - 1/3²)(1 - 1/4²)......(1 - 1/k²)(1 - 1/(k + 1)²)

Now, from step 2, we get

So, (1 - 1/(k + 1)²)((k + 1)/2k)

or, ((k + 1)/2k)((k² + 1 + 2k - 1)/(k + 1)²) 

or, k(k + 2)/2k(k + 1)

or, (k + 2)/2(k + 1)

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 36. Prove that (2n)!/2²ⁿ(n!)² ≤ 1/√(3n + 1) is true for all n ∈ N.

Solution:

Let, P(n) = (2n)!/2²ⁿ(n!)² ≤ 1/√(3n + 1) 

Step 1:

Now, let us check P(n) for n = 1.

P(1) = 1/2 ≤ 1/√3 + 1 = 1/2

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m.

P(m) = (2m)!/2^2m(m!)² ≤ 1/√(3m + 1) 

Step 3:

Now, we have to prove for P(m + 1) is true. i.e. 

P(m + 1) = (2m + 2)!/2^2m+2(m!)² ≤ 1/√(3m + 4)  

Now,

P(m + 1) = ((2m + 1)(2m + 1)(2m)!)/(2m².2²(m + 1)²(m!)²)

or, (2m + 2)!/2{2m + 2}((m + 1)!)² = ((2m)!/2^2m) × (2m + 1)(m + 2)/2²(m + 1)²

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ (2m + 1)/2(m + 1)√(3m + 1)

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √((2m + 1)²/4(m + 1)²(3m + 1))

(2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √((2m + 1)²/4(m + 1)²(3m + 1))

or, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ √(12m³ + 28m² + 19m + 4)/(12m³ + 28m² + 20m + 4)(3m + 4)

now, (12m³ + 28m² + 19m + 4)/(12m³ + 28m² + 20m + 4) < 1

so, (2m + 2)!/2{2m + 2}((m + 1)!)² ≤ 1/√(3m + 4)

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 37. Prove that 1 + 1/4 + 1/9 + 1/16 + .....+ 1/n² < 2 - 1/n for all n > 2, n ∈ N.

Solution:

Let P(n) = 1 + 1/4 + 1/9 + 1/16 + .....+ 1/n² < 2 - 1/n for all n > 2, n ∈ N

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1/2² < 2 - 1/2

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m.

P(m) = 1 + 1/4 + 1/9 + 1/16 +.....+ 1/m² < 2 - 1/m

Step 3:

Now, we have to prove for P(m + 1) is true. i.e.,

From step 2 we have, 1 + 1/4 + 1/9 + 1/16 +.....+ 1/m² < 2 - 1/m 

Now, adding 1/(m + 1)² to both the sides, we get

= 1 + 1/4 + 1/9 + 1/16 +.....+ 1/m² + 1/(m + 1)² < 2 - 1/m + 1/(m + 1)²

or, (m + 1)² > m + 1

or, 1/(m + 1)² < 1/(m + 1)

or, 1/m - 1/(m + 1)² < 1/(m + 1)

So, P(m + 1) < 2 - 1/(m + 1)

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 38. Prove that x^2n-1 + y^2n-1 is divisible by x + y.

Solution:

Let, P(n) be   x^2n-1 + y^2n-1

Step 1:

Now, let us check P(n) for n = 1.

P(1) = x + y  

So, P(1) is divisible by x + y.

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m.

P(m) = x^2m-1 + y^2m-1= λ(x + y)         ........... (i)

Step 3:

Now, we have to prove for P(m + 1) is true. 

i.e. ,  P(m + 1) = x^2m+1 + y^2m+1

= x^2m+1 + y^2m+1 - x^2m-1. y²  +  x^2m-1 . y²

= x^2m-1(x² - y²) + y²(x^2m-1+y^2m-1)

= (x + y)(x^2m-1 (x - y) + λy²)

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 39. Prove that sinx + sin3x + ...+ sin(2n - 1)x = sin²nx/sinx is true for all n ∈ N.

Solution:

Let, P(n) =  sinx + sin3x + ...+ sin(2n - 1)x = sin²nx/sinx

Step 1:

Now, let us check P(n) for n = 1.

P(1) = sin x = sin² x / sin x = sin x

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = m.

P(m) =  sinx + sin3x + ...+ sin(2m - 1)x = sin²mx/sinx

Step 3:

Now, we have to prove for P(m + 1) is true.

i.e., 

P(m + 1) = sinx + sin3x +...+ sin(2m - 1)x + sin(2m + 1)x = (sin²mx/sinx) + sin(2m+1)x

now,

P(m + 1) = {(sin²mx + sinx[sin(mx) + cos(m + 1)x + sin(m + 1)x + cos(mx)])}/sinx

= {(sin²mx + 2sinx.cosx.cos(mx) - sin²x.sin²mx + cos²mx.sin²x)}/sinx

= {(sin²mx(1 - sin²x) + 2sinx.cosx.cos(mx) + cos²mx.sin²x)}/sinx

= {(sin²mx.cos²x + 2sinx.cosx.cos(mx) + cos²mx.sin²x)}/sinx

= {(sin(mx).cosx + cos(mx).sinx)²}/sinx

= {(sin(m + 1)x)²}/sinx  

So, P (n) is true for n = m + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 40. Prove that cosα + cos(α +β) +cos(α +2β)+....+cos(α +(n-1)β)=\frac{(cos[α +(\frac{n-1}{2})β] sin(\frac{nβ}{2}))}{sin(\frac{β}{2})}  is true for all n ∈ N.

Solution:

Let, P(n) = cosα + cos(α +β) +cos(α +2β)+....+cos(α +(n-1)β)=\frac{(cos[α +(\frac{n-1}{2})β] sin(\frac{nβ}{2}))}{sin(\frac{β}{2})} 

Step 1:

Now, let us check P(n) for n = 1.

L.H.S = cos [α + (1 - 1)β] = cos α

R.H.S =\frac{(cos[α +(\frac{1-1}{2})β] sin(\frac{β}{2}))}{sin(\frac{β}{2})} = cosα

So, P(1) is true as LHS and RHS are equal.

Step 2:

Let us consider P (n) be the true for n = k.

P(k) = cosα + cos(α +β) +cos(α +2β)+....+cos(α +(k-1)β)=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2}))}{sin(\frac{β}{2})} 

Step 3:

Now, we have to prove for P(k + 1) is true.

So, adding cos(α + kβ) both sides of P(k), we get

P(k+1)= cosα + cos(α +β) +cos(α +2β)+....+cos(α +(k-1)β) +cos(α +kβ)=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2}))}{sin(\frac{β}{2})}+cos(α +kβ)

=\frac{(cos[α +(\frac{k-1}{2})β] sin(\frac{kβ}{2})+cos(α +kβ)sin(\frac{β}{2})}{sin(\frac{β}{2})}

=\frac{(-sin(α-\frac{β}{2}) + sin(α+kβ+\frac{β}{2}))}{2sin(\frac{β}{2})}

=\frac{2cos(\frac{(2α +kβ)}{2})sin(\frac{(kβ+β)}{2})}{2sin(\frac{β}{2})}

=\frac{cos(α +\frac{kβ}{2})sin(\frac{(k+1)β}{2})}{sin(\frac{β}{2})}

Now,

RHS= \frac{(cos[α +(\frac{kβ}{2})] sin(\frac{(k+1)β}{2}))}{sin(\frac{β}{2})}

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 41. Prove that 1/(n+1) + 1/(n+2) +.....+ 1/2n > 13/24 for all natural numbers, n > 1.

Solution:-

Let,P(n) = 1/(n+1) + 1/(n+2) +.....+ 1/2n > 13/24

Step 1:

Now, let us check P(n) for n = 2.

P(2) = 1/(2 + 1) + 1/(2 + 2) = 1/3 + 1/4 = 7/12 > 13/24

So, P(2) is true.

Step 2:

Let us consider P (n) be the true for n = k, So, P(k) is

P(k) = 1/(k + 1) + 1/(k + 2) +.....+ 1/2k > 13/24

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

So, P(k + 1) = 1/(k + 2) + 1/(k + 3) +.....+ 1/2k + 1/2(k + 1)

Here, as LHS = RHS

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 42. Given a₁ = 1/2(a₀ + A/a₀), a₂ = 1/2(a₁ + A/a₁) and a_n+1 = 1/2(a_n + A/a_n), a, A > 0

To prove: \frac{(a_n - √A)}{(a_n + √A)}  = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(n-1)}}

Solution:

Let, P(n) = \frac{(a_n - √A)}{(a_n + √A)}  = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(n-1)}}

Step 1:

Now, let us check P(n) for n=1.

LHS = (a₁ - √A) / (a₁ + √A) 

RHS = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(1-1)}}

= (a₁ - √A) / (a₁ + √A)

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k, So, P(k) is

So, P(k) = \frac{(a_k - √A)}{(a_k + √A)}  = (\frac{ a_1 - √A}{a_1 + √A})^{2^{(k-1)}}

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

So, P(k + 1) = LHS = (a_k+1 - √A) / (a_k+1 + √A)

=  (1/2(a_k + A/a_k) - √A) /  (1/2(a_k + A/a_k) + √A) 

= (1/2(a_k²+ A - 2a_k√A)/a_k ) / (1/2(a_k²+ A + 2a_k√A)/a_k )

=  (a_k+1 - √A)² / (a_k+1 + √A)²

= [ [\frac{a_1 - √A}{a_1 + √A}]^{2^{k-1}}]^2 

= [\frac{a_1 - √A}{a_1 + √A}]^{2^k}

here, as LHS = RHS

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 43. Let P(n) be the statement: 2ⁿ ≥ 3n. If P(r) is true then show that P(r + 1)is true, Do you conclude that P(n) is true for all n ∈ N?.

Solution:

Let P(n) = 2ⁿ ≥ 3n

Step 1:

Now, let us check P(n) for n = 1.

L.H.S = 2

R.H.S = 3

As L.H.S < R.H.S

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = r, So, P(r) is 2^r ≥ 3r

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

So, P(k + 1) = 2^r+1 = 2.2^r

For, x > 3, 2x > x + 3

So, 2.2^r > 2^r+3 for r > 1

or,2^r+1 > 2^r+3 for r > 1

or, 2^r+1 > 3r +3 for r > 1

or, 2^r+1 > 3(r + 1) for r >1

So, P (n) is true for n = r + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 44. Show by the principle of Mathematical Induction that the sum Sn of the n terms of the series 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² + 7² +... is given by 

S_n= \begin{cases}   \frac{n(n+1)^2}{2}, if \ n \ is \ even, \\   \frac{n^2(n+1)}{2}, if \ n \ is \ odd \end{cases}

Solution:

Let, P(n) = S_n = 1² + 2 × 2² + 3² + 2 × 4² + 5² + 2 × 6² + 7² +... = \begin{cases}   \frac{n(n+1)^2}{2}, if \ n \ is \ even, \\   \frac{n^2(n+1)}{2}, if \ n \ is \ odd \end{cases}

Step 1:

Now, let us check P(n) for n = 1.

LHS = 1 = RHS

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k, So, P(k) is

P(k) = 1² + 2.2² + 3² + 2.4² + 5² = \begin{cases}   \frac{k(k+1)^2}{2}, if \ k \ is \ even, \\   \frac{k^2(k+1)}{2}, if \ k \ is \ odd \end{cases}

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

Case1: When k is odd, then (k + 1) is even

P(k + 1) = LHS = 1² + 2.2² + 3² + 2.4² + 5².... k² + 2.(k + 1)²

= k²(k + 1)/2 + 2.(k + 1)²

= {(k²(k + 1) + 4(k + 1)²)}/2

= (k + 1)(k + 2)²/2

now, RHS = (k + 1)(k + 1 + 1)²/2

= (k + 1)(k + 2)²/2

So, it is true for n = k + 1, when k is odd.

Case 2: When k is even, then (k + 1) is odd

P(k + 1) = LHS = 1² + 2.2² + 3² + 2.4² + 5²....+ 2. k² + (k + 1)²

= k(k + 1)²/2 + (k + 1)²

= (k(k + 1)² + 2.(k + 1)²)/2

= (k + 1)²(k + 2)/2

RHS = (k + 1)²(k + 1 + 1)/2

= (k + 1)²(k + 2)/2

now, LHS = RHS.

So, it is true for n=k+1, when k is even.

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 45.  Prove that the number of subsets of a set containing n distinct elements is 2ⁿ for all n ∈ N.

Solution:

Let, P(n): The number of subsets of a set containing n distinct elements = 2ⁿ, for all n ∈ N.

Step 1:

Now, let us check P(n) for n = 1.

LHS = As, the subsets of the set containing only 1 element are:

Φ and the set itself

i.e. the number of subsets of a set containing only element=2

R.H.S = 2¹ = 2

now, LHS = RHS

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

So, P(k) is The number of subsets of a set containing k distinct elements = 2^k

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

P(k + 1) = Let A = {a₁, a₂, a₃, a₄,..…, a_k, b} so that A has (k + 1) elements.

Now, the subset t of A can be divided into two collections such that 

First contains subsets of A which don’t have b in them and

The second contains subsets of A which do have b in them.

So, First collection: {}, {a₁}, {a₁, a₂}, {a₁, a₂, a₃},…,{a₁, a₂, a₃, a₄,…, a_k} and

and the second collection: {b}, {a₁, b}, {a₁, a₂, b}, {a₁, a₂, a₃, b},…,{a₁, a₂, a₃, a₄,…,a_k, b}

It can be clearly seen that:

The number of subsets of A in first collection

= The number of subsets of set with k elements i.e. {a₁, a₂, a₃, a₄,…, a_k} = 2k

Also, it follows that the second collection must have

the same number of the subsets as that of the first = 2^k

So the total number of subsets of A = 2^k + 2^k = 2^k+1

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 46. A sequence a₁, a₂, a₃,..... is defined by letting a₁ = 3 and a_k = 7a_k-1 for all numbers natural numbers k ≥ 2. Show that a_n = 3.7^n-1 for all n ∈ N.

Solution:

Let P(n) be a_n = 3.7^n-1 for all n ∈ N

Step 1: 

Now, let us check P(n) for n = 1.

so, a₁ = 3.7^1-1 = 3

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P(k) = a_k = 3.7^k-1

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

P(k + 1) = a_k+1 = 7.a_k

= 7.3.7^k-1

= 3.7^k-1+1

= 3.7^(k+1)-1

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 47. A sequence x₁, x₂, x₃,..... is defined by letting x₁ = 2 and x_k = x_k-1 /k for all numbers natural numbers k, k ≥ 2. Show that x_n = 2/n! for all n ∈ N.

Solution:

Let, P(n) be x_n = 2/n!  for all n ∈ N.

Step 1:

Now, let us check P(n) for n = 1.

P(1) = x₁ = 2/1! = 2

So, P(1) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P(k) = x_k = 2/k! 

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

so,  x_k+1 = 2/(k + 1)! 

or, 2/(k + 1).k!

or, 2/(k + 1)!

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 48. A sequence x₀, x₁, x₂, x₃, ……. is defined by letting x₀ = 5 and x_k = 4 + x_{k -1} for all natural number k. Show that x_n = 5 + 4_n for all n ∈ N using mathematical induction.

Solution:- 

Let, P(n) = x_n = 5 + 4n for all n ∈ N

Step 1:

Now, let us check P(n) for n = 0.

P(0) = 5 + 4(0) = 5

So, P(0) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P(k) = x_k = 5 + 4k  

Step 3:

Now, we have to prove for P(k + 1) is true. When P(k) is true.

so, P(k + 1) = x_k+1 = 4 + x_{k+1 -1}

= 4 + x_k

= 4 + 5 + 4k

= 5 + 4(k + 1) 

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ∈ N

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ∈ N.

Question 49. Using principle of mathematical induction prove that

√n < 1/√1 + 1/√2+ 1/√3 +.....+1/√n for all natural numbers n ≥2.

Solution:

Let, P(n) = √n < 1/√1 + 1/√2+ 1/√3 +.....+1/√n for all n ≥ 2.

Step 1:

Now, let us check P(n) for n=2.

P(2) = √2 < 1 + 1/√2

or, 1.41 < 1 + 0.707 = 1.707

So, P(2) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P(k) = √k < 1/√1 + 1/√2+ 1/√3 +.....+1/√k

Step 3:

Now, we have to prove for P(k+1) is true. When P(k) is true.

Now, LHS = √(k + 1)

Now, RHS = 1/√1 + 1/√2+ 1/√3 +.....+1/√k + 1/√(k + 1)

or, k/√{(k + 1)} < √k

or, k + 1/√{(k + 1)} - 1/√(k + 1) < √k

or, √(k + 1) - 1/√(k + 1)< √k

or, √(k + 1) < √k + 1/√(k + 1)

so, LHS < RHS.

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ≥ 2

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ≥ 2.

Question 50. The distributive law from algebra states that for all real numbers, a₁ and a₂, we have c (a₁ + a₂) = ca₁ + ca₂. Use this law and mathematical induction to prove that, for all natural numbers, n ≥ 2, if c, a₁, a₂,...,a_n are any real numbers, then c (a₁+a₂+…+a_n) = ca₁+ca₂+…+ca_n.

Solution:

Let, P(n) = c (a₁+a₂+…+a_n) = ca₁+ca₂+…+ca_n, for all natural numbers, n ≥ 2. 

Step 1: 

Now, let us check P(n) for n=2.

LHS = c(a₁ + a₂)

RHS = c a₁ + ca₂

So, P(2) is true.

Step 2:

Let us consider P (n) be the true for n = k,

P(k) =  c(a₁+a₂+…+a_k) = ca₁+ca₂+…+ca_k

Step 3:

Now, we have to prove for P(k+1) is true. When P(k) is true.

LHS = c(a₁+a₂+…+a_k + a_k+1)

= c[(a₁+a₂+…+a_k) + a_k+1]

= c(a₁+a₂+…+a_k) + ca_k+1

= ca₁+ca₂+…+ca_k + ca_k+1

= RHS

So, P (n) is true for n = k + 1

i.e., P (n) is true for all n ≥ 2

Hence, by principle of Mathematical Induction (PMI), P (n) is true for all n ≥ 2.

Summary

Exercise 12.2 | Set 3 of Chapter 12 in RD Sharma's Class 11 mathematics textbook offers a rigorous and advanced application of Mathematical Induction. Through a carefully curated set of complex problems involving higher-order polynomial sums, divisibility properties, intricate inequalities, and challenging number theory concepts, students are pushed to apply inductive reasoning at an expert level. This set not only reinforces the fundamental principles of mathematical induction but also significantly expands students' problem-solving capabilities, encouraging them to approach sophisticated mathematical statements with confidence and analytical precision. By mastering these problems, students develop a profound appreciation for the power and versatility of mathematical induction, preparing them for advanced mathematical concepts in higher studies, competitive examinations, and fields such as computer science, engineering, and pure mathematics.